Problem 11 - Thomas Calculus

Question

In Exercises \(11-14,\) find the arc length parameter along the curve from the point where \(t=0\) by evaluating the integral $$ s=\int_{0}^{t}|\mathbf{v}(\tau)| d \tau $$ from Equation \((3) .\) Then find the length of the indicated portion of the curve. $$ \mathbf{r}(t)=(4 \cos t) \mathbf{i}+(4 \sin t) \mathbf{j}+3 t \mathbf{k}, \quad 0 \leq t \leq \pi / 2 $$

Step-by-Step Solution

Answer

The length of the curve is \(\frac{5\pi}{2}\).

Step 1: Find the Velocity Vector

Calculate the velocity vector \(\mathbf{v}(t)\), which is the derivative of the position vector \(\mathbf{r}(t)\). The position vector is given by \(\mathbf{r}(t) = (4\cos t) \mathbf{i} + (4\sin t) \mathbf{j} + 3t \mathbf{k}\). The derivative is computed component-wise:\[\mathbf{v}(t) = \frac{d}{dt}(4\cos t) \mathbf{i} + \frac{d}{dt}(4\sin t) \mathbf{j} + \frac{d}{dt}(3t) \mathbf{k}\]\[= (-4\sin t) \mathbf{i} + (4\cos t) \mathbf{j} + 3 \mathbf{k}\]Thus, the velocity vector is \(\mathbf{v}(t) = (-4\sin t) \mathbf{i} + (4\cos t) \mathbf{j} + 3\mathbf{k}\).

Step 2: Calculate Magnitude of Velocity Vector

Find \(|\mathbf{v}(t)|\), the magnitude of the velocity vector \(\mathbf{v}(t)\):\[|\mathbf{v}(t)| = \sqrt{(-4\sin t)^2 + (4\cos t)^2 + 3^2}\]\[= \sqrt{16\sin^2 t + 16\cos^2 t + 9}\]Using the Pythagorean identity, \(\sin^2 t + \cos^2 t = 1\), simplify:\[= \sqrt{16(\sin^2 t + \cos^2 t) + 9} = \sqrt{25} = 5\]Thus, the magnitude is \(|\mathbf{v}(t)| = 5\).

Step 3: Evaluate the Integral for Arc Length

The arc length parameter \(s\) from \(t=0\) to \(t=\frac{\pi}{2}\) is given by the integral:\[s = \int_{0}^{t} |\mathbf{v}(\tau)| \, d\tau\]Since \(|\mathbf{v}(\tau)| = 5\) is constant, the integral becomes:\[s = \int_{0}^{t} 5 \, d\tau = 5\tau \bigg|_{0}^{t} = 5t\]Thus, the arc length parameter is \(s = 5t\).

Step 4: Calculate Arc Length for Given Interval

To find the length of the curve from \(t=0\) to \(t=\frac{\pi}{2}\), substitute \(t=\frac{\pi}{2}\) into \(s=5t\):\[s = 5 \times \frac{\pi}{2} = \frac{5\pi}{2}\]The length of the indicated portion of the curve is \(\frac{5\pi}{2}\).

Key Concepts

Velocity Vector

To understand the arc length parameter of a curve, start by determining the velocity vector of the curve. The velocity vector is derived from the position vector, the function that defines the position of a point along a curve as a function of time. In this exercise, the position vector is given by:\[ \mathbf{r}(t) = (4 \cos t) \mathbf{i} + (4 \sin t) \mathbf{j} + 3t \mathbf{k} \]To find the velocity vector, you take the derivative of the position vector with respect to the parameter \(t\). This involves differentiating each component of \( \mathbf{r}(t) \):- The derivative of \(4 \cos t\) is \(-4 \sin t\)- The derivative of \(4 \sin t\) is \(4 \cos t\)- The derivative of \(3t\) is a constant, \(3\)Putting these together gives the velocity vector:\[ \mathbf{v}(t) = (-4 \sin t) \mathbf{i} + (4 \cos t) \mathbf{j} + 3 \mathbf{k} \]This vector describes the speed and direction of movement along the curve at any point \(t\).

Pythagorean Identity

A key simplification was possible in this exercise due to the Pythagorean Identity, which states \( \sin^2 t + \cos^2 t = 1 \).This identity is fundamental in trigonometry and helps simplify expressions involving squares of sines and cosines.Let's take a closer look at how it was applied in finding the magnitude of the velocity vector, \(|\mathbf{v}(t)|\). The magnitude is calculated using the formula:\[ |\mathbf{v}(t)| = \sqrt{(-4 \sin t)^2 + (4 \cos t)^2 + 3^2} \]Applying the Pythagorean Identity simplifies it as follows:- \((-4 \sin t)^2\) becomes \(16 \sin^2 t\)- \((4 \cos t)^2\) becomes \(16 \cos^2 t\)- Adding these, we apply the identity:\[ 16 \sin^2 t + 16 \cos^2 t = 16 \]This means we have:\[ |\mathbf{v}(t)| = \sqrt{16 + 9} = \sqrt{25} = 5 \]Hence, using this identity simplifies the process dramatically. It's crucial in many length-related calculations involving trigonometric functions.

Magnitude of a Vector

Understanding the magnitude of a vector is key when dealing with arc length. The magnitude of a vector gives us the length or size of the vector, essentially how fast the point on the curve is moving. For the velocity vector \( \mathbf{v}(t) = (-4 \sin t) \mathbf{i} + (4 \cos t) \mathbf{j} + 3 \mathbf{k} \), you'll want to determine its magnitude to proceed with arc length calculations. The formula for the magnitude \(|\mathbf{v}(t)|\) is:\[ |\mathbf{v}(t)| = \sqrt{(-4 \sin t)^2 + (4 \cos t)^2 + 3^2} \]Breaking it down, this entails:- Squaring each component of the velocity vector- Summing these squares- Taking the square root of this sumAfter simplifying with the Pythagorean Identity, the expression simplifies to:\[ |\mathbf{v}(t)| = 5 \]This constant magnitude makes calculations straightforward. It directly impacts the calculation of the arc length.

Integral for Arc Length

To find the arc length parameter, the integral of the magnitude of the velocity vector is evaluated over the interval of interest. In this problem, the integral takes the form:\[ s = \int_0^t |\mathbf{v}(\tau)| \, d\tau \]The integral calculates the arc length from \(t = 0\) to any point \(t\), effectively measuring distance traveled along the curve. With a constant magnitude, the calculation simplifies significantly:- Given \(|\mathbf{v}(\tau)| = 5\), a constant, the integral becomes:\[ s = \int_0^t 5 \, d\tau = 5\tau \bigg|_0^t = 5t \]Evaluating this at \(t = \frac{\pi}{2}\) yields the arc length for the specified portion of the curve:\[ s = 5 \cdot \frac{\pi}{2} = \frac{5\pi}{2} \]This result indicates the precise length of the curve's segment from \(t=0\) to \(t=\frac{\pi}{2}\). Calculating arc length through integration is a powerful tool for understanding geometric properties of a curve.